Melody

The sum of 2 cubics is  ..  (something you need to memorize)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

So if you have a denominator of  \((a+b)\) where a or b have cube roots, and you multiply it by  \((a^2-ab+b^2)\) then you will end up with a rational denominator.

\(\frac{1}{\sqrt[3]{2}+\sqrt{16}}=\frac{1}{\sqrt[3]{2}+4}\)

let

\(a= 2^{1/3}\qquad    and\qquad b=4\\so\\ a^2=2^{2/3}\qquad    and\qquad b^2=16\\ ab=4*2^{1/3}\)

\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

\(\frac{1}{\sqrt[3]{2}+4}\times \frac{2^{2/3}-4*2^{(1/3)}+4^2}{2^{2/3}-4*2^{(1/3)}+4^2}\\~\\ =\frac{2^{2/3}-4*2^{(1/3)}+4^2}{2+64}\\~\\ =\frac{\sqrt[3]{4}-4\sqrt[3]{2}+16}{66}\\~\\\)

Here is a video that might help you understand

You are a member because you have a username.  You are not a 'guest'.

Most members have the free version.  I think this just means that since you didn't pay anything, that you see some ads. 

I just realized.

Thie repost is for a question that you had only just posted.

Please do not expect instant answers. 

As a member, you will tend to get priority answering anyway.  

Reposting so quickly will get you a bad reputation which will not help you in the longer term.

A guest has answered your question correctly.

Here is your graph abcd...

I have not thought about it very hard but I also cannot see, offhand, how this graph shows anything.   

I also ask, what do you want to do with it?

\(\sqrt{160-64\sqrt3}\\ =4\sqrt{10-4\sqrt3}\)

Let the numbers be a and b

\(10^5 \le a,b \le 10^6-1\\ so\\ 10^{10} \le ab \le (10^6-1)^2\\ 10^{10} \le ab \le 9.99998*10^{11}\\ \frac{10^{10}}{10} \le\frac{ ab}{10}\qquad \qquad \frac{ ab}{100} \le \frac{9.99998*10^{11}}{100}\\ \text{I only care about the smallest one so ...}\\ 10^{9} \le\frac{ ab}{ \textcolor{red}{10}} \qquad \qquad \frac{ ab}{100} \le 9.99998*10^{9}\)

\(\frac{ab}{10 }\ge \text{10 digits}\qquad \frac{ab}{100}< 11\;\;digits\)

The smallest common divisor must be between 10 and 100 inclusive.

The smallest common divisor is bigger or equal to 10  and 10 has  2 digits.

ok, so what did you figure out?

thanks for showing me what you have done.

Your graph looks interesting but I can't say more.

My memory of complex numbers is rudimentary at best. 

Yes that is right.

You have, or can have   (x+y+z) four times.  Put them altogether.   

example

if I have   6(x+3)+m(x+3)

then I have 6 lots of x+3 and I add m lots of x+3   I end up with  6+m lots of x+3

so

6(x+3)+m(x+3) = (6+m) (x+3)